Significant figures in one of the Sears and Zemansky book problems?

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SUMMARY

The forum discussion focuses on calculating instantaneous acceleration using the formula v = 60 m/s + (0.50 m/s³) t². The user attempts to find the instantaneous acceleration at t=1.0s by evaluating the change in velocity (delta v) over small time intervals (delta t). The correct instantaneous acceleration converges to 1.0 m/s² as delta t approaches zero, emphasizing the importance of maintaining significant figures throughout calculations to avoid losing precision.

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FountainDew
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Homework Statement


Average acceleration:
- The instantaneous velocity at any time can be calculated with the given formula: v = 60 m/s + (0.50 m/s^3) t^2
- Find the instantaneous acceleration at t=1.0s, by taking (delta)t to be 0.1s, then 0.01s, and then 0.001s.

Homework Equations


a = (delta)v / (delta)t

The Attempt at a Solution


@t = 1.0s
v = 60.5 m/s

@t=1.1s
v = 60 m/s + (0.50 m/s^3)(1.1 s)^2
v = 60.61 m/s

a = (delta)v / (delta)t = (60.61 m/s - 60.5 m/s) / (0.1 s) = 1.1 m/s^2

etc... (repeated for 0.01s and 0.001s, not important to my question)

4. The actual solution
@t = 1.0s
v = 60.5 m/s

@t=1.1s
v = 60 m/s + (0.50 m/s^3)(1.1 s)^2
v = 60.605 m/s

a = (delta)v / (delta)t = (60.605 m/s - 60.5 m/s) / (0.1 s) = 1.05 m/s^2
etc...

5. The Question
If 1.1 squared is 1.21, then if it is multiplied with 0.50 it should equal 0.605. But according to the rule of multiplication of significant figures, the result is equal to the least amount of sig fig's. So it should equal 0.61? This is significant to me, because as you go down the question, sig figs matter more and more and more!

What am I doing wrong?
 
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I am not sure if you have had calculus yet, but the instantaneous acceleration at t=1.0 seconds is 1.0 m/s^2. As you take your delta t to smaller and smaller values, the solution converges to a = 1.0. If you round off to sig figures, it's like throwing the baby out with the bathwater.
 
PhanthomJay said:
I am not sure if you have had calculus yet, but the instantaneous acceleration at t=1.0 seconds is 1.0 m/s^2. As you take your delta t to smaller and smaller values, the solution converges to a = 1.0. If you round off to sig figures, it's like throwing the baby out with the bathwater.

Thanks for the quick reply! And I didn't do so well in calculus, but I'm going to keep trying! The problem made more sense once I started keeping the numbers after the decimal intact, I could see it slowly moving to 1.0 but never reaching it. By rounding it early, I lost the entire sense of the problem! The baby and the bathwater analogy helped me understand, thanks for that imagery. :)
 

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